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Direct Comparison Test

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What is the Direct Comparison Test ???

The Formula

Let $0 \le a_{n}\leq b_{n} $ for all n
If $\sum\limits_{n=1}^{\infty}b_{n} $ converges, then $\sum\limits_{n=1}^{\infty}a_{n} $ also converges
If $\sum\limits_{n=1}^{\infty}a_{n} $ diverges, then $\sum\limits_{n=1}^{\infty}b_{n} $ also diverges

Examples

If $\lim\limits_{n\rightarrow \infty}\frac{a_{n}}{b_{n}}=\rho , 0<\rho<\infty $,
then both converge or both diverge .
If $ \lim\limits_{n\rightarrow \infty}\frac{a_{n}}{b_{n}}=0 \text{, and } \sum b_{n} $ converges
then $\sum a_{n} $ converges
If $ \lim\limits_{n\rightarrow \infty}\frac{a_{n}}{b_{n}}=\infty \text{, and } \sum b_{n} $ diverges
Then $\sum a_{n} $ diverges.
To apply the Comparison Test to a series $\sum c_{n} $ you need to find another series $\sum u_{n} $ that will allow for a useful comparison.
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When should you use this test ???

  • Case 1: If you suspect that $\sum c_{n} $ converges, you should look for a convergent series $\sum u_{n} $ that is larger then $\sum c_{n} $.
  • Case 2: If you suspect that $\sum c_{n} $ diverges, you should look for a divergent series $\sum u_{n} $ that is smaller then $\sum c_{n} $
However. The Direct Comparison Test is most often used to compare series that are similar to a P-series or Geometric Series. Furthermore, when searching for comparison, we often take the given series and make a new series with only the highest degree of n in the numerator and denominator

Examples

Series Comparison Relation
$\sum\limits_{n=1}^{\infty}\frac{5}{2+3n+n^4} $ $\sum\limits_{n=1}^{\infty}\frac{5}{n^4} $ $\frac{5}{2+3n+n^4} \leq \frac{5}{n^4} $
Conclusion: Since $\sum\limits_{n=1}^{\infty}\frac{5}{n^4} $ converges so does $\sum\limits_{n=1}^{\infty}\frac{5}{2+3n+n^4} $
Series Comparison Relation
$ \sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n^2-1}} $ $\sum\limits_{n=1}^{\infty}\frac{1}{n} $ $\frac{1}{\sqrt{n^2-1}} \geq \frac{1}{n} $
Conclusion: Since $ \sum\limits_{n=1}^{\infty}\frac{1}{n} $ diverges so does $ \sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n^2-1}} $
Series Comparison Relation
$ \sum\limits_{n=1}^{\infty}\frac{1}{3^{n}+2} $ $ \sum\limits_{n=1}^{\infty}\frac{1}{3^{n}} $ $ \frac{1}{3^{n}+2} \leq \frac{1}{3^{n}} $
Conclusion: Since $ \sum\limits_{n=1}^{\infty}\frac{1}{3^{n}} $ converges so does $ \sum\limits_{n=1}^{\infty}\frac{1}{3^{n}+2} $
Series Comparison Relation
$ \sum\limits_{n=1}^{\infty}\frac{\sqrt{n}}{n^2+1} $ $ \sum\limits_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}} $ $ \frac{\sqrt{n}}{n^2+1} \leq \frac{1}{n^{\frac{3}{2}}} $
Conclusion: Since $ \sum\limits_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}} $ converges so does $\sum\limits_{n=1}^{\infty}\frac{\sqrt{n}}{n^2+1} $
Series Comparison Relation
$ \sum\limits_{n=1}^{\infty}\frac{1+ln(n)}{n} $ $ \sum\limits_{n=1}^{\infty}\frac{1}{n} $ $ \frac{1+ln(n)}{n} \geq \frac{1}{n} $
Conclusion: Since $\sum\limits_{n=1}^{\infty}\frac{1}{n} $ diverges so does $\sum\limits_{n=1}^{\infty}\frac{1+ln(n)}{n} $
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